Question:

A cell is constructed by Cl\(_2\)/Cl\(^-\)(aq) and a standard hydrogen electrode half-cells. The standard potential of the complete cell is 1.38 V and \( \left(\frac{\partial E^0}{\partial T}\right)_P = -1.24 \, \text{mV K}^{-1} \). The \( \Delta S^0_{\text{reaction}} \) (in J K\(^{-1}\) mol\(^{-1}\)) for the following cell reaction is .................. \[ \text{H}_2 \text{(g)} + \text{Cl}_2 \text{(g)} \rightarrow 2 \text{H}^+ \text{(aq)} + 2 \text{Cl}^-\text{(aq)} \]

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To find the entropy change for a reaction, use the relationship between cell potential and temperature, and apply the equation \( \Delta S^0_{\text{reaction}} = -n F \left(\frac{\partial E^0}{\partial T}\right)_P \).
Updated On: Sep 8, 2025
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Solution and Explanation

To calculate the change in entropy \( \Delta S^0_{\text{reaction}} \), we can use the relation between the standard electrode potential and the entropy change: \[ \Delta G^0_{\text{reaction}} = -n F E^0_{\text{cell}} \] where:
- \( n = 2 \) (the number of electrons involved in the reaction),
- \( F = 96480 \, \text{C mol}^{-1} \) (Faraday constant),
- \( E^0_{\text{cell}} = 1.38 \, \text{V} \) (standard cell potential).
From the relationship between Gibbs free energy and entropy: \[ \Delta G^0_{\text{reaction}} = \Delta H^0_{\text{reaction}} - T \Delta S^0_{\text{reaction}} \] At constant pressure, \( \Delta H^0_{\text{reaction}} = 0 \) (since the reaction involves no change in enthalpy at standard conditions), so: \[ \Delta S^0_{\text{reaction}} = - \frac{\Delta G^0_{\text{reaction}}}{T} \] Now, using the derivative of the standard potential with respect to temperature \( \left(\frac{\partial E^0}{\partial T}\right)_P = -1.24 \, \text{mV K}^{-1} = -1.24 \times 10^{-3} \, \text{V K}^{-1} \), the entropy change is: \[ \Delta S^0_{\text{reaction}} = -n F \left(\frac{\partial E^0}{\partial T}\right)_P \] Substituting the values: \[ \Delta S^0_{\text{reaction}} = -2 \times 96480 \times (-1.24 \times 10^{-3}) \] \[ \Delta S^0_{\text{reaction}} = 239.6 \, \text{J K}^{-1} \, \text{mol}^{-1} \] Final Answer: \[ \boxed{239.6} \]
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